r-value > Probability

I have two wrestlers who are going to arm-wrestle. I want to try to predict who will win based on their previous fights against other opponents.

This is the information that I know:

The outcome of each wrestler’s last 5 fights (win, lose, draw).

I know that the probability of a win is 0.4, the probability of a loss of 0.4 and the probability of a draw is 0.2 (these are true in the population of arm-wrestlers as a whole).

If we give 2 points for a win, 1 point for a draw and 0 points for a loss, then I can work out the Average and Standard Deviation for each wrestler’s last 5 fights.

Using these I can work out the Variance, Degrees of Freedom and the Test Statistic t. I can then work out the r-value (correlation co-efficient). The r-value gives me an indication of who will win the fight, or if the fight will be a draw.

But the r-value does not tell me: What is the probability of Wrestler A winning, what is the probability of wrestler B winning, and what is the probability of a draw?

Is it possible from the information I have to work out these probabilities? Can I turn an r-value into a probability (p) value?

I know this r-value shows that Wrestler A has a slight advantage in this fight. But is there a way of working out the exact probabilities of each of the three outcomes (Wrestler A wins, Wrestler B wins, draw)?

Re: r-value > Probability

Originally Posted by cmarkc

I know this r-value shows that Wrestler A has a slight advantage in this fight. But is there a way of working out the exact probabilities of each of the three outcomes (Wrestler A wins, Wrestler B wins, draw)?

Thanks. Marc

Probably not. You could use something like an Elo rating system and the assumption that each wrestler previous opponents represent average opposition.

Then the Elo system would give a probability of win based on the two players rating difference (or rather the expected score of the game which is not quite the same thing because of the possibility of draws)